Fundamental Theorem of Calculus Examples

Example 1. Calculate the derivative of the function [Tex]F(x) = \int_{1.5}^x \sqrt{(t^2 + 3t)} dt          [/Tex]at x = 3.

Solution: 

Given that [Tex]F(x) = \int_{1.5}^x \sqrt{(t^2 + 3t)} dt [/Tex]

So, using first FTC, we get

F'(x) =  [Tex]= \frac{d}{dx} [ \int_{1.5}^x f(t) dt          [/Tex]= √(x² + 3x)

Therefore,

F'(3) = √(3² + 3.3) = √(9 + 9) = √(2.9) = 3√2.

Example 2. Calculate the derivative of the function [Tex]F(x) = \int_{\frac{-\pi}{2}}^x \sqrt{(sin t + cos t)}dt         [/Tex] at x = π/2.

Solution: 

Given that [Tex]F(x) = \int_{\frac{-\pi}{2}}^x \sqrt{(sin t + cos t)}dt[/Tex]

So, using first FTC, we get

F'(x) = [Tex]\frac{d}{dx} [\int_{\frac{-\pi}{2}}^x f(t) dt         [/Tex] = f(x)

Hence,

F'(x) = √(sin x + cos x)

Therefore,

F'(π/2) = √(sin π/2 + cos π/2) = √(1 + 0) = 1.

Example 3. Find the derivative of [Tex]F(x) = \int_{3}^{x^2} \frac{dt}{t^2}         [/Tex].

Solution: 

Given that [Tex]F(x) = \int_{3}^{x^2} \frac{dt}{t^2}[/Tex]

So, let x² = u. 

Now, consider a new function,

G(u) = ∫₃^u dt/t² 

G'(u) = 1/u²

Since, F(x) = G(x²)

F'(x) = G'(x²).2x

F'(x) = (1/x⁴).2x

F'(x) = 2/x³.

Example 4. Find the derivative of the [Tex]F(x) = \int_{0}^{x^2} \sqrt{(1+t^3)}dt         [/Tex].

Solution: 

Given that [Tex]F(x) = \int_{0}^{x^2} \sqrt{(1+t^3)}dt[/Tex]

So, let x² = u. 

Now, consider a new function,

G(u) = ∫₀^u √(1 + t³) dt

Using first FTC, we have

G'(u) = √(1 + u³)

As F(x) = G(x²)

F'(x) = G'(x²).2x

F'(x) = 2x √(1 + (x²)³)

F'(x) = 2x √(1 + x⁶).

Example 5. Find the derivative of the function [Tex]F(x) = \int_{1}^{x^2} (t^2+t) dt         [/Tex].

Solution: 

Given that [Tex]F(x) = \int_{1}^{x^2} (t^2+t) dt[/Tex]

So, let x² = u. 

Now, considering new function,

G(u) =  ∫₁^u (t² + t) dt.

Using first FTC, we get

G'(u) = u² + u

Since, F(x) = G(x²)

F'(x) = G'(x²).2x

F'(x) = (x⁴ + x).2x

F'(x) = 2x⁵ + 2x².

Example 6. Evaluate the integral [Tex]\int_{x}^{x^2} t dt         [/Tex] .

Solution: 

Given that [Tex]\int_{x}^{x^2} t dt[/Tex]

Now we find the anti-derivative of t

∫t dt = t² + C

Now using the second FTC, we get

∫_a^b f(x) dx = F(b) – F(a)

[Tex]\int_{x}^{x^2} t dt         [/Tex] = F(x²) – F(x)

[Tex]\int_{x}^{x^2} t dt         [/Tex] = ((t²)²) – (t²)

[Tex]\int_{x}^{x^2} t dt         [/Tex] = (t⁴) – (t²)

[Tex]\int_{x}^{x^2} t dt         [/Tex] = t²((t²) – 1)

Example 7. Evaluate the integral [Tex]\int_{1}^{2} x dx         [/Tex] .

Solution: 

Given that I = [Tex]\int_{1}^{2} x dx[/Tex]

Now using the second FTC, we get

∫_a^b f(x) dx = F(b) – F(a)

[Tex]\int_{1}^{2} x dx         [/Tex] = F(2) – F(1)

Now we find the anti-derivative of x

∫x dx = x² + C

[Tex]\int_{1}^{2} x dx         [/Tex] = (2²) – 1

= 4 – 1

= 3

Example 8: Evaluate the integral ∫₀² (x² – x) dx.

Solution: 

Given I = ∫₀² (x² – x) dx

Using second FTC, we get

∫_a^b f(x) dx = F(b) – F(a)

∫₀² (x² – x) dx = F(2) – F(0)

Now we find the anti-derivative of x

∫₀² (x² – x) dx = x³/3 – x²/2

So, 

= [x³/3 – x²/2]₀²

= [8/3 – 4/2]

= [8/3 – 2]

= [8/3 – 6/3]

= 2/3.

Example 9: Evaluate the integral  F(x) = ∫₀^x e^t + e^-t dt.

Solution:

Given that F(x) = ∫₀^x e^t + e^-t dt

Using the first FTC, we get

F'(x) = e^x + e^-x.

Example 10: Find the integral of [Tex]\int_{0}^{ln 2} t.e^{-t} dt       [/Tex].

Solution:

Given I = [Tex]\int_{0}^{ln 2} t.e^{-t} dt[/Tex]

so we can also write as

I = [Tex]\int_{0}^{ln 2} t.d(e^{-t})[/Tex]

Now apply integration by parts, we get

u = t, dv = d(e^-t)

So, du = 1, v = e^-t

So, the integral is

I = –[Tex]\int_{0}^{ln 2} t.d(e^{-t})[/Tex]

=[Tex] [(te^{-t})|^{ln2}_0 –  \int_0^{ln2} e^{-t} dt][/Tex]

= [Tex] [(te^{-t})|^{ln2}_0 +  \int_0^{ln2} e^{-t} dt][/Tex]

=[Tex] -(te^{-t})|^{ln2}_0 -(e^{-t})|^{ln2}_ 0[/Tex]

= [Tex]- [e^{-t}(t + 1)]|^{ln2}_0[/Tex]

= -e^-ln2(ln2 + 1) + e⁰ .1

= 1/2(ln e – ln 2)

= 1/2 ln e/2

Fundamental Theorem of Calculus is the basic theorem that is widely used for defining a relation between integrating a function of differentiating a function. The fundamental theorem of calculus is widely useful for solving various differential and integral problems and making the solution easy for students.

This is widely used in the fields of physics, engineering, medicine, economics, biology, space exploration, statistics, pharmacology, and many more.

Before learning about the fundamental theorem of calculus let’s first learn about calculus and others. In this article, we will learn about calculus, area function, the fundamental theorem of calculus, and others.